Can you prove/disprove $\int_{-\infty}^{\infty}{\frac{x!}{(x-t)!(t)!}dt} = 2^x$

Question

I was studying something about binomial coefficients then I thought that whether I can expand the formulae $\sum_{i=0}^n {\binom{n}{i}}= 2^n$ to integrals. So I tried plugging in some values in calculator, and here are my results:-

★Sometimes the ratio is above 1 $$\frac {\int_{-170}^{170}{\frac{(3.9)!}{(3.9-t)!(t)!}dt}} {2^{3.9}} = 1.0000008068$$

★Sometimes the ratio is below 1 $$\frac {\int_{-170}^{170}{\frac{(3.1)!}{(3.1-t)!(t)!}dt}} {2^{3.1}} = 0.999999987055$$

★It is also worth noting that the ratio isn't close to 1 when we plug in negative values, may be because the function in integration diverges at both ends but I think it should converge when we go towards infinity $$\frac {\int_{-100}^{100}{\frac{(-1.8)!}{(-1.8-t)!(t)!}dt}} {2^{-1.8}} = 221.136809313$$

Answer 1

Computing $$I_x=\frac{x!}{2^x}\int_{-\infty}^{+\infty}\frac{dt}{(x-t)!\,\,t!}$$ I had the following results (convergence problems for $x<3$) $$\left( \begin{array}{cc} 3 & 1.00000593804190 \\ 4 & 0.99999769880684 \\ 5 & 1.00000027889204 \\ 6 & 0.99999972800274 \\ 7 & 0.99999999215455 \\ 8 & 0.99999990105455 \\ 9 & 0.99999997983950 \\ 10 & 1.00000000122939 \\ 11 & 1.00000000353610 \\ 12 & 0.99999999425838 \\ 13 & 0.99999998263212 \\ 14 & 0.99999999013611 \\ 15 & 0.99999999858660 \\ 16 & 0.99999999334312 \\ 17 & 1.00000001664105 \\ 18 & 1.00000000019262 \\ 19 & 0.99999999948438 \\ 20 & 0.99999999938931 \\ 21 & 1.00000000450905 \\ 22 & 1.00000000000426 \\ 23 & 1.00000000000573 \\ 24 & 0.99999999999661 \\ 25 & 0.99999999999690 \\ 26 & 1.00000000002458 \\ 27 & 1.00000000000656 \\ 28 & 0.99999999990548 \\ 29 & 1.00000000000046 \\ 30 & 1.00000000000054 \\ 31 & 1.00000000048450 \\ 32 & 1.00000000000213 \\ 33 & 1.00000000000252 \\ 34 & 1.00000000000235 \\ 35 & 1.00000000000202 \\ 36 & 1.00000000000168 \\ 37 & 1.00000000000137 \\ 38 & 1.00000000000110 \\ 39 & 1.00000000000088 \\ 40 & 1.00000000000075 \\ 41 & 1.00000000000064 \\ 42 & 1.00000000000045 \\ 43 & 1.00000000000020 \\ 44 & 0.99999999999875 \\ 45 & 1.00000000000045 \\ 46 & 1.00000000000065 \\ 47 & 1.00000000000002 \\ 48 & 0.99999999999871 \\ 49 & 0.99999999999836 \\ 50 & 1.00000000000075 \end{array} \right)$$

Answer 2

We have:

$$\int_{-\infty}^\infty\binom0t~\mathrm dt=\int_{-\infty}^\infty\frac{\sin(\pi t)}{\pi t}~\mathrm dt=1$$

and

$$\int_{-\infty}^\infty\binom xt~\mathrm dt=\int_{-\infty}^\infty\binom{x-1}{t-1}+\binom{x-1}t~\mathrm dt=2\int_{-\infty}^\infty\binom{x-1}t~\mathrm dt$$

which proves this is indeed equivalent to $2^x$ for integer $x$ where it converges. Further we know that this is at the very least a piecewise exponential function, depending on what happens for $x\in(0,1)$.

Answer 3

I think this is a rather well-known result, although my quick googling does not reveal any references. So here is my own proof of the following general result:

Theorem. Let $\alpha$ be a complex number away from negative integers, and denote by

$$ \binom{\alpha}{z} := \frac{\alpha!}{z!(\alpha-z)!} = \frac{\Gamma(\alpha+1)}{\Gamma(z+1)\Gamma(\alpha-z+1)} $$

the extended binomial coefficient. Then for $\Re(\alpha) > 0$ and $x \in \mathbb{R}$, we have

$$ \binom{\alpha}{x} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{\sin \pi(x-n)}{\pi(x-n)} \tag{1} $$ and $$ \int_{\mathbb{R}}\binom{\alpha}{x} e^{-2\pi i \xi x} \, \mathrm{d}x = \begin{cases} \big(1 + e^{-2\pi i \xi}\big)^{\alpha} & |\xi| < \frac{1}{2} \\ 0 & \text{otherwise}. \end{cases} \tag{2} $$

Remark. Your question corresponds to $\xi = 0$ case.

Proof. Define $I(\alpha, x)$ by

$$ I(\alpha, x) = \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{2\pi i \xi}\big)^{\alpha} e^{-2\pi i x \xi} \, \mathrm{d}\xi. $$

We begin by proving that $I(\alpha, x)$ equals the right-hand side of $\text{(1)}$. Indeed, by the extended binomial theorem,

\begin{align*} I(\alpha, x) &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \biggl( \sum_{n=0}^{\infty} \binom{\alpha}{n} e^{2\pi n i \xi} \biggr)e^{-2\pi i x \xi} \, \mathrm{d}\xi = \sum_{n=0}^{\infty} \binom{\alpha}{n} \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{2\pi (n-x) i \xi} \, \mathrm{d}\xi \\ &= \sum_{n=0}^{\infty} \binom{\alpha}{n} \left[ \frac{e^{2\pi (n-x) i \xi}}{2\pi (n-x)} \right]_{-\frac{1}{2}}^{\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{\sin \pi(x-n)}{\pi(x-n)} \end{align*}

This calculation is fully justified once we establish the validity of switching the order of the integration and summation. This is accomplished either by the Weierstrass $M$-test or by the Fubini's Theorem, together with the estimate:

$$ \left|\binom{\alpha}{n}\right| = O\left(\frac{1}{n^{1+\Re(\alpha)}}\right) \qquad\text{as}\quad n\to\infty, $$

which itself follows from the Stirling's approximation:

\begin{align*} \left|\binom{\alpha}{n}\right| &= \frac{1}{n!}\left| \prod_{k=1}^{n} (\alpha - k + 1) \right| = \frac{1}{n!}\left| \prod_{k=1}^{n} (k - \alpha - 1) \right| = \frac{1}{n!}\left| \frac{\Gamma(n-\alpha)}{\Gamma(\alpha)} \right| \\ &\leq C \left| \frac{n^{-1/2} \left(\frac{n-\alpha}{e}\right)^{n-\alpha}}{n^{1/2} \left(\frac{n}{e}\right)^n} \right| \leq \frac{C}{n^{1+\Re(\alpha)}}. \end{align*}

Next, we show the that

$$ I(\alpha, x) = \binom{\alpha}{x} \tag{3} $$

holds. Once this is established, $\text{(1)}$ is immediate and $\text{(2)}$ is a consequence of the Fourier Inversion Theorem.

To prove $\text{(3)}$, note that both sides of $\text{(3)}$ define entire functions in $x$. So, by the principle of analytic continuation, it suffices to verify the equality on a set having an accumulation point.

Now choose the principal branch cut for the complex logarithm and make the substitution $z = e^{2\pi i \xi}$. If $\mathcal{C}$ parametrizes the unit circle minus the principal branch cut in the counterclockwise orientation, then

\begin{align*} I(\alpha, x) &= \frac{1}{2\pi i} \int_{\mathcal{C}} (1+z)^{\alpha}z^{-x-1} \, \mathrm{d}z \end{align*}

For the purpose of the proof, we restrict ourselves to the case $x < 0$. Then the integrand is analytic on $\mathbb{C}\setminus(-\infty, 0]$. By invoking the keyhole contour,

\begin{align*} I(\alpha, x) &= \frac{1}{2\pi i} \lim_{\epsilon \to 0^+} \biggl( \int_{-1-\epsilon i}^{-\epsilon i} (1+z)^{\alpha}z^{-x-1} \, \mathrm{d}z + \int_{\epsilon}^{-1+\epsilon i} (1+z)^{\alpha}z^{-x-1} \, \mathrm{d}z \biggr) \\ &= \frac{1}{2\pi i} \biggl( \int_{0}^{1} (1-t)^{\alpha}e^{-i\pi(-x-1)} t^{-x-1} \, \mathrm{d}t - \int_{0}^{1} (1-t)^{\alpha}e^{i\pi(-x-1)}t^{-x-1} \, \mathrm{d}t \biggr) \\ &= \frac{\sin(-\pi x)}{\pi} \int_{0}^{1} (1-t)^{\alpha} t^{-x-1} \, \mathrm{d}t. \end{align*}

Finally, using the beta function identity and Euler's reflection formula, the above simplifies to

$$ I(\alpha, x) = \frac{1}{\Gamma(-x)\Gamma(x+1)} \cdot \frac{\Gamma(\alpha+1)\Gamma(-x)}{\Gamma(\alpha-x+1)} = \binom{\alpha}{x}. $$

Therefore $\text{(3)}$ holds for $ x < 0$, and thus for all of $x \in \mathbb{C}$, completing the proof.